Generalization of Hensel lemma: �nding of roots

of p-adic Lipschitz functions(joint talk with Andrei Khrennikov)

Dr. Ekaterina Yurova Axelsson

Linnaeus University, Sweden

September 8, 2015

Outline

I De�nitions

I Classical Hensel's lifting lemma

I Generalization of "Hll" for 1-Lipschitz functions

I Generalization of "Hll" for pα-Lipschitz functions

De�nitions: p-adic numbers

I For any prime p ≥ 2 the p-adic norm | · |p is de�ned in thefollowing way. For every nonzero integer n let ordp(n) be the highestpower of p which divides n, i.e. n ≡ 0 (mod pordp(n)), n 6≡ 0(mod pordp(n)+1). Then we de�ne |n|p = p−ordp(n), |0|p = 0. Forrationals n

m ∈ Q we set | nm |p = p−ordp(n)+ordp(m).

I The completion of Q with respect to the p-adic metricρp(x , y) = |x − y |p is called the �eld of p-adic numbers Qp. Thenorm satis�es the strong triangle inequality |x ± y |p ≤ max |x |p; |y |pwhere equality holds if |x |p 6= |y |p.

I The set Zp = {x ∈ Qp : |x |p ≤ 1} is called the set of p-adicintegers.

I Every x ∈ Zp can be expanded in canonical form, i.e. in aconvergent by p-adic norm series:

x = x0 + px1 + . . .+ pkxk + . . . , xk ∈ {0, 1, . . . , p − 1}, k ≥ 0.

I The ball of radius p−r with center a is the setBp−r (a) = {x ∈ Zp : |x − a|p ≤ p−r} = a + prZp.

De�nitions: p-adic Lipschitz functions

I Consider functions f : Zp → Zp, which satisfy the Lipschitzcondition with constant pα, α ≥ 0 (pα-Lipschitz functions).

I Recall that f : Zp → Zp is pα-Lipschitz function if

|f (x)− f (y)|p ≤ pα|x − y |p, for all x , y ∈ Zp.

This condition is equivalent to that from x ≡ y (mod pk) followsf (x) ≡ f (y) (mod p k−α) for all k ≥ 1 + α.

I A class of 1-Lipschitz functions take a special place (i.e. α = 0).For all k ≥ 1 a 1-Lipschitz transformation f : Zp → Zp the reducedmapping modulo pk is fmodpk : Z/pkZ→ Z/pkZ, z 7→ f (z)(mod pk). Mapping fmodpk is a well-de�ned (the fmodpk does notdepend on the choice of representative in the ball z + pkZp).

Hensel's lifting lemma

The main tool for �nding the roots of p-adic functions that map the ringof p-adic integers into itself, is a classical result - Hensel's liftinglemma.

Theorem (Hensel's lifting lemma for p-adic case)Let f (x) ∈ Zp[x ] be a polynomial with integer p-adic coe�cients and

f ′(x) ∈ Zp[x ] be its formal derivative. Suppose that a ∈ Zp is a p-adicinteger such that f (a) ≡ 0 (mod p) and f ′(a) 6≡ 0 (mod p).Then there exists a unique p-adic integer a ∈ Zp such that f (a) = 0 and

a ≡ a (mod p).

Classical Hensel's lifting lemma

Example. Find the solutions of x3 + x2 + 29 ≡ 0 (mod 25).Solution. Let f (x) = x3 + x2 + 29. We see (by inspection) that solutionsof f (x) ≡ 0 (mod 5) have x ≡ 3 (mod 5). Because f

′(x) = 3x2 + 2x

and f′(3) = 33 ≡ 3 6≡ 0 (mod 5), Hensel's lemma tells us that there is a

unique solution modulo 25 of the form 3 + 5t, where t ≡ −f ′(3)( f (3)5

)

(mod 5). Note that f ′(3) = 3 = 2, because 2 is inverse to 3 mod 5. Also

note that f (3)5

= 655

= 13. It follows that t ≡ −2 · 13 ≡ 4 (mod 5). Weconclude that x ≡ 3 + 5 · 4 = 23 is the unique solution of f (x) ≡ 0(mod 25).

Hensel's lifting lemma

Hensel's lifting lemma (Hll) for p-adic functions characterized by thefollowing circumstances:

1. under restrictions on the function f (value of the derivative f at agiven point) Hll gives the criterion of the existence of the root ofp-adic function f on the set of p-adic integers. By solving a �nitenumber of congruences modulo p we can determine whether f hasthe root in Zp;

2. Hll provides an algorithm for �nding the approximate value (inp-adic metric) of the root of f with a given accuracy. This problemis reduced to solving linear congruences modulo p at each step ofthe iterative procedure. Solvability of the corresponding linearcongruences is determined by the restrictions that are imposed onthe derivative f ′;

3. Hll is applicable for p-adic functions de�ned by polynomials withinteger p-adic coe�cients. Even if the function f is de�ned bypolynomial, but not with the p-adic integer coe�cients, the"formal" use of Hensel's lifting lemma to �nd the approximate valueof the root leads to false results.

Question arises:

I How could we �nd roots of p-adic function that maps p-adicintegers into itself and is not de�ned by a polynomial with integercoe�cients, for example, if f is not di�erentiable at all?

I We solved this problem for the p-adic functions that satisfy theLipschitz condition with constant pα, α ≥ 0.

Results: 1-Lipschitz functions

General criterion for the existence of the root of 1-Lipschitz functions:Theorem 1.Let f : Zp → Zp be 1-Lipschitz function. The function f has a root ifand only if the congruences fmodpk (x) ≡ 0 (mod pk) are solvable for anyk ≥ 1, where fmodpk : Z/pkZ→ Z/pkZ, fmodpk (z) = f (z) (mod pk).

It follows that, in general, for 1-Lipschitz function f by solving a �nitenumber of congruences modpk it is impossible to determine whether thefunction f has root in Zp, as it is done with the use of classical Hensel'slifting lemma.

Roots of 1-Lipschitz functions. Example

Let p = 5 and f : Z5 → Z5 such that

f (x) = f (x0 + 5x1 + . . .+ 5kxk . . .) = x + 5N(−xN + x2N + 3),

where N is a �xed positive integer. Note that f is 1-Lipschitz function.We need to solve a �nite number of congruences of the formfmod5k (x) ≡ 0 (mod 5k), k ≤ S . Let N > S . It is clear thatfmod5k (0) ≡ 0 (mod 5k), k ≤ S . If a direct analogy with the Hensel'slifting lemma is true, then we can assume that x = 0 is the root of thefunction f . However, the congruence f (x) ≡ 0 (mod 5N+1) has nosolutions.Indeed, suppose that h ∈ {0, . . . , 5N+1 − 1} be a solution of thiscongruence and h = h + 5NhN . Because f (h) ≡ f (h) ≡ 0 (mod 5N),then h = 0 and f (h) ≡ 5N(h2N + 3) ≡ 0 (mod 5N+1) or h2N + 3 ≡ 0(mod 5). But the last congruence has no solutions in Z/5Z. FromTheorem it follows that the function f has no roots at all.Thus, by choosing N su�ciently large, we cannot by solving a �nitenumber of congruences modulo 5k , k ≤ S determine the existence of theroot of the function f .

Roots of 1-Lipschitz functions

In this regard, we need some restrictions on the 1-Lipschitz function.

I We would like to know when one can determine whether or not thefunction has a root and �nd the approximate value of this root inthe p-adic metric with the given accuracy by solving a �nite numberof congruences.

I In other words, under what restrictions on the 1-Lipschitzfunction do we get an analogue of Hensel's lifting lemma?

I The following theorem gives such restrictions in terms of theproperties of the van der Put coe�cients. The next Theoremgeneralizes Hensel's lifting lemma for 1-Lipschitz functions.

De�nitions: van der Put series

Given a continuous function f : Zp → Zp, there exists a unique sequenceB0,B1,B2, . . . of p-adic integers such that for all x ∈ Zp

f (x) =∞∑

m=0

Bmχ(m, x). (0.1)

Let m = m0 + . . .+ mn−2pn−2 + mn−1p

n−1 be a base-p expansion for m,i.e.,mj ∈ {0, . . . , p − 1}, j = 0, 1, . . . , n − 1 and mn−1 6= 0, then

Bm =

{f (m)− f (m −mn−1p

n−1), if m ≥ p;

f (m), otherwise.

De�nitions: van der Put series

The characteristic function χ(m, x) is given by

χ(m, x) =

{1, if |x −m|p ≤ p−n

0, otherwise

where n = 1 if m = 0; and n is uniquely de�ned by the inequalitypn−1 ≤ m ≤ pn − 1 otherwise.The number n is just the number of digits in a base-p expansion ofm ∈ N0. Then⌊logp m

⌋= (the number of digits in a base-p expansion for m)− 1,

therefore n =⌊logp m

⌋+ 1 for all m ∈ N0 and

⌊logp 0

⌋= 0 (bαc for a

real α denotes the integral part of α).

De�nitions: p-adic Lipschitz functions

1-Lipschitz functions f : Zp → Zp in terms of the van der Put series weredescribed by W. Schikhof. But we follow notations from papers byAnashin, Khrennikov, Yurova in convenience for further study.The function f is 1-Lipschitz if and only if it can be represented as

f (x) =∞∑

m=0

pblogp mcbmχ(m, x) (0.2)

for suitable bm ∈ Zp, m ≥ 0.So for every a ∈ {0, 1, . . . , pk − 1}, k ≥ 1 we set functionsψa : {0, . . . , p − 1} → {0, . . . , p − 1} de�ned by the relations:

ψa(i) =

{ba+pk i (modp), i 6= 0;

0, i = 0.(0.3)

Generalization of Hensel's lifting lemma for 1-Lipschitz

functions

Theorem 2. Let f : Zp → Zp be a 1-Lipschitz function represented via

van der Put series f (x) =∑∞

m=0 pblogp mcbmχ(m, x).If

1. for some natural number R there existsh ∈ {0, 1, . . . , pR − 1} such that f (h) ≡ 0 (mod pR) and

2. for any m ≥ R, where m ≡ h (mod pR) functionsψm : {0, . . . , p − 1} → Z/pZ,

ψm(i) =

{bm+i·p1+blogp mc(modp), i 6= 0;

0, i = 0.

are bijective (i.e. bm+i·p1+blogp mc(modp), i = 1, 2, . . . , p − 1 are all

nonzero residues modulo p),

then there exists a unique p-adic integer h ∈ Zp such that f (h) = 0 andh ≡ h (mod pR).

Roots of 1-Lipschitz functions

I From Theorem 2 follows the "classical" Hensel's lifting lemma.

I Criterion of the existance of the root in Theorem 2 does not requirethat functions are polynomials (as in Hensel's lifting lemma).Moreover, this Theorem does not require that the functions aredi�erentiable.

I The proof is constructive, i.e. there is an algorithm to construct anapproximate value of the root of f with given p-adic accuracy(similar to proof of Hensel's lemma). Di�erence: at each step wesolve nonlinear congruence modulo p.

I Solution of such nonlinear congruences reduces to calculation of pvalues of the function by suitable modulo pk .

Example

Let p = 5 and

f (x) = f (x0 + 5x1 + . . .+ 5k · xk + . . .) = −3 +∞∑k=0

5k(1 + 5k2)x3k .

We show that f has a unique root and �nd its approximation within 5−4

in 5-adic metric. Note that f is non-di�erentiable at any point from Z5.Indeed, for any k > 1 and x ∈ {0, . . . , 5k − 1}

f (x +5kh)− f (x) = 5k(1+5k2)h3 ≡ 5kh3 (mod 5k+1), h ∈ {1, 2, 3, 4}.

This means that we cannot use Hensel's lifting lemma for �nding theroots of f but we can use Theorem 2 instead.

Example

Let us �nd the coe�cients of Bm, m ≥ 5 with the aid of the van der Putseries of the function f . Letm = m + 5k · i , m ∈ {0, . . . , 5k − 1}, i ∈ {1, 2, 3, 4}, k ≥ 1, then

Bm = f (m + 5k · i)− f (m) = 5k(1 + 5k2) · i3.

In this notation, the functions ψm, m ≥ 5 can be represented as

ψm(i) ≡ i3 (mod 5)

Because gcd(3, 4) = 1, then ψm, m ≥ 5 are bijective on Z/5Z (i.e. thesecond condition of Theorem 2 holds).Since x ≡ x0 = 2 (mod 5) is the only solution of the congruencef (x) ≡ x30 − 3 ≡ 0 (mod 5), then by Theorem 2 function f has a uniqueroot in Z5, namely, h, and h ≡ 2 (mod 5).

Example

Now we �nd an approximation of the root h to within 5−4. Ifh = h0 + 5h1 + . . .+ 5khk + . . . , hk ∈ {0, 1, 2, 3, 4}, then to �nd suchapproximation it is necessary to determine the values h0, h1, h2, h3. Letus use an iterative procedure [this paper].Value h0 has already been de�ned, i.e. h0 = 2. Leth(s) = h0 + . . .+ 5shs , s ∈ {0, 1, 2, 3}.Value h1 ∈ {0, 1, 2, 3, 4} we determine from the condition

f (h(0) + 5 · h1) = f (2 + 5 · h1) ≡ 0 (mod 52).

By computing f (2 + 5 · i) (mod 52), i ∈ {0, 1, 2, 3, 4}, we get h1 = 4 andh(1) = 22.Value h2 ∈ {0, 1, 2, 3, 4} we determine from the condition

f (h(1) + 5 · h2) = f (22 + 52 · h1) ≡ 0 (mod 53).

By computing f (22 + 5 · i) (mod 53), i ∈ {0, 1, 2, 3, 4}, we get h2 = 2and h(2) = 72.

Example

Value h3 ∈ {0, 1, 2, 3, 4} we determine from the condition

f (h(2) + 53 · h3) = f (72 + 53 · h3) ≡ 0 (mod 54).

By computing f (72 + 53 · i) (mod 54), i ∈ {0, 1, 2, 3, 4}, we get h3 = 2and h(3) = 322.Thus, the approximate value of the root of f to within 5−4 is 322, i.e.|h − 322|5 ≤ 5−4.

Roots of pα-Lipschitz functions

I Then we generalize Hensel's lifting lemma for the case ofpα-Lipschitz functions.

I The problem of �nding the roots and their approximations forpα-Lipschitz functions (α > 0) is reduced to solving thecorresponding problem for 1-Lipschitz sub-functions.

I Such possibility follows from the following Theorem.

Generalization of Hensel's lifting lemma for pα-Lipschitzfunctions

Theorem 3. (representation) Let f : Zp → Zp be a pα-Lipschitzfunction (α ≥ 1) represented as

f (a + pαx) =

pα−1∑i=0

Ii (a) · fi (x), a ∈ {0, . . . , pα − 1}, (0.4)

where all the functions fi (x) : Zp → Zp, i ∈ {0, . . . , pα − 1} satisfyLipschitz condition with constant 1 and

Ii (a) =

{1, if a = i ;

0, otherwise.

Criterion of the existance of the root

Let f : Zp → Zp be a pα-Lipschitz function. 1-Lipschitz sub-functionsfi , i ∈ {0, . . . , pα − 1} are represented with the aid of the van der Put

series fi (x) =∑∞

m=0 pblogp mcbi, m · χ(m, x). If

1. for some natural number R there existsh ∈ {0, 1, . . . , pR+α − 1} such that f (h) ≡ 0 (mod pR);

2. for any m ≥ R, where m ≡ h−spα (mod pR), s ≡ h (mod pα)

functionsψs, m : {0, . . . , p − 1} → Z/pZ,

ψs, m(t) =

{bs, m+t·p1+blogp mc(modp), t 6= 0;

0, t = 0.

are bijective (i.e. bs, m+t·p1+blogp mc(modp), t = 1, 2, . . . , p − 1 are all

nonzero residues modulo p),

then there exists a unique p-adic integer h ∈ Zp such that f (h) = 0 andh ≡ h (mod pR+α).

Roots of pα-Lipschitz functions

I To check existance of the root of the pα-Lipschitz function f , weneed to check existance of the roots of 1-Lipschitz sub-functionsfi (x), i ∈ {0, . . . , pα − 1};

I Algorithm for construction of approximate root of pα-Lipschitzfunction is also presented.

Roots of pα-Lipschitz functions. ExampleLet p = 5 and

f (x0 + 5x1 + . . .+ 5kxk + . . .) =∞∑k=1

5k(1 + 5x0)xx0k .

We �nd approximation of all roots of the function f (x)− 2 to within5−3. We represent the function f (x)− 2 in the form of the statement ofthe Theorem 3, we obtain (using the notation x = x1 + 5x2 + . . .)

f (x0 + 5x)− 2 =

= I0(x0)f0(x) + I1(x0)f1(x) + I2(x0)f2(x)+

+ I3(x0)f3(x) + I4(x0)f4(x),

where

f0(x) = −2 +∞∑k=1

5k−1 = −9

4; f1(x) = −2 + 6

∞∑k=1

5k−1xk ;

f2(x) = −2 + 11

∞∑k=1

5k−1x2k ; f3(x) = −2 + 16

∞∑k=1

5k−1x3k ;

f4(x) = −2 + 21

∞∑k=1

5k−1x4k

Roots of pα-Lipschitz functions. Example

I It is easy to see that all the functions fi satisfy the Lipschitzcondition with constant 1. For each i = 0, 1, 2, 3, 4 we �nd solutionsof the congruence fi (x0) ≡ 0 (mod 5). We get f1(2) ≡ 0(mod 5); f3(3) ≡ 0 (mod 5); and for i = 0, 2, 4 appropriatecongruences have no solutions (this means that the numbers of theform 5x , 2 + 5x and 4 + 5x are not the roots of the functionf (x)− 2).

I For the functions f1 and f3 let us verify conditions of the Theorem 3.For this we �nd the van der Put coe�cients of the functions f1 andf3. Let m = m + 5kt, m ∈ {0, . . . , 5k − 1}, t ∈ {1, 2, 3, 4}, then

B1, m = f1(m)−f1(m) = 5k(6·t), B3, m = f3(m)−f3(m) = 5k(16·t3).

Roots of pα-Lipschitz functions. Example

I Accordingly, functions ψ1, m and ψ3, m take the form

ψ1, m(t) ≡ t (mod 5), ψ3, m(t) ≡ t3 (mod 5)

and, therefore, functions ψ1, m and ψ3, m are bijective. This meansthat the function f (x)− 2 has exactly two roots. Using an iterativeprocedure [from this paper] and initial approximations for f1 and f3,we �nd solutions of congruences f1(x) ≡ 0 (mod 53) and f3(x) ≡ 0(mod 53).

I As a result, we obtain

f1(2 + 5 · 3 + 52 · 1) = f1(42) ≡ 0 (mod 53)

f3(3 + 5 · 4 + 52 · 2) = f3(73) ≡ 0 (mod 53).

Thus, the function f (x)− 2 has two roots.

I The approximate value of these roots in 5-adic metric to within 5−3

are 1 + 5 · 42 = 211 and 3 + 5 · 73 = 368, respectively.

Conclusions

I We generalized Hensel's lifting lemma for a wider class of functions,namely, the class of pα-Lipschitz functions.

I We presented an algorithm for construction of approximate root ofsuch functions with given p-adic accuracy.

I "Generalization of Hensel lemma: �nding of roots of p-adicLipschitz functions" by E. Yurova Axelsson and A. Khrennikov,Journal of Number Theory, Elsevier, volume 158, p. 217-233(January 2016)